Question 55 of 130

A square shaped hole of side$l=\dfrac{a}{2}$ is carved out at a distance $d=\dfrac{a}{2}$ from the centre '$O$' of a uniform circular disk of radius a. If the distance of the centre of mass of the remaining portion from $O$ is $-\dfrac{a}{x}$, value of $X$ (to the nearest integer) is :

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Answer: 23

${X}_{a}=\dfrac{A\cdot X-A\cdot X1}{A-{A}_{1}}=\dfrac{{\pi a}^{2}\times 0-\dfrac{{a}^{2}}{4}\times \dfrac{a}{2}}{{\pi a}^{2}-\dfrac{{a}^{2}}{4}}$
$=\dfrac{-{a}^{3}/8}{\left(\pi -\dfrac{1}{4}\right){a}^{2}}=\dfrac{-a}{2(4\pi -1)}=\dfrac{-a}{8\pi -2}=-\dfrac{a}{23}$
$x=23$